Tag Info

New answers tagged


$$(17\sqrt2-24)(17\sqrt2+24)=17^2\times2-24^2=289\times2-576=2$$ so $$0<\sqrt2-{24\over17}={2\over17(17\sqrt2+24)}\dot={1\over\sqrt217^2}$$ where $a\dot=b$ means $a$ is very close to $b$; in particular, $0<\sqrt2-{24\over17}<{1\over17^2}$. But the convergents to the continued fraction for $\sqrt2$ are $1/1,3/2,7/5,17/12,41/29,\dots$ so $24/17$ is ...


The consecutive convergents are alternately greater than and less than the value of $b$. Then we have that $${1\over q_nq_{n+1}}=\left|{p_{n+1}q_n-p_nq_{n+1}\over q_nq_{n+1}}\right|=\left|{p_{n+1}\over q_{n+1}}-{p_n\over q_n}\right|=\left|{p_{n+1}\over q_{n+1}}-b\right|+\left|{p_{n}\over q_{n}}-b\right|$$ The last equality follows because they one is ...


The simultaneous version of Dirichlet's theorem asserts that you can find infinitely many $a_n, b_n$ and $q_n$ such that $$ \left| \alpha - \frac{a_n}{q_n} \right| < \frac{1}{q_n^{3/2}} \; \mbox{ and } \; \left| \beta - \frac{b_n}{q_n} \right| < \frac{1}{q_n^{3/2}}. $$ For almost all pairs of real numbers (including, via a theorem of Schmidt, pairs of ...


Hint: $2 < \sqrt{6} < 3$ so it starts $2 + 1/\ldots$. If $\sqrt{6} = 2 + 1/x$ then $x = \dfrac{1}{-2+\sqrt{6}} = \dfrac{-2-\sqrt{6}}{4 - 6} = \ldots$


The non-closed-formedness of that answer is that the size of the given expression depends on the size of the problem. That's quintessentially not a closed form, just as much as something with, say, $\sum_{k=1}^n$. The fact that it's finite for any particular invocation is completely beside the point.

Top 50 recent answers are included